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PhysicsModern PhysicsPhotoelectric effectHard2 minPYQ_2020

PhysicsHardstatement

When radiation of wavelength $A$ is used to illuminate a metallic surface, the stopping potential is $V$ . When the same surface is illuminated with radiation of wavelength $3 A$ , the stopping potential is $\frac{V}{4}$ . If the threshold wavelength for the metallic surface is $nλ$ then value of $n$ will be :

Answer:

Solution:

$\frac{h c}{λ} = ϕ + e V$ ...........(1)

$\frac{h c}{3 λ} = ϕ + \frac{e V}{4}$ ---------(2)

from (1) & (2)

$\frac{h c}{λ} (1 - \frac{1}{3}) = \frac{3}{4} e v$

$\frac{hc}{λ} \frac{2}{3} = \frac{3}{4} eV$

$eV = \frac{8}{9} \frac{hc}{λ}$

$\frac{h c}{λ} = ϕ + \frac{8}{9} \frac{h c}{λ}$

$ϕ = \frac{hc}{9 λ} = \frac{hc}{λ_{th}}$

$λ_{t h} = 9 λ$

$∴ k = 9$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Photoelectric effect

⏱ 2mℹ️ Source: PYQ_2020

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